What is the integral from 2 to 9 of 4/(8x-9) ?

The integral from 2 to 9 of 4/(8x-9) is ln(3)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 2 to 9 of 4/(8x-9)

\int_{2}^{9} \frac{4}{8 x - 9} d x

ln (3)

Solution steps

\int_{2}^{9} \frac{4}{8 x - 9} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \int_{2}^{9} \frac{1}{8 x - 9} d x

Apply u-substitution

: \int_{7}^{63} \frac{1}{8 u} d u

= 4 \cdot \int_{7}^{63} \frac{1}{8 u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \frac{1}{8} \cdot \int_{7}^{63} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 4 \cdot \frac{1}{8} [ln ∣ u ∣]_{7}^{63}

Simplify

4 \cdot \frac{1}{8} [ln ∣ u ∣]_{7}^{63} : \frac{1}{2} [ln ∣ u ∣]_{7}^{63}

= \frac{1}{2} [ln ∣ u ∣]_{7}^{63}

Compute the boundaries:

ln (63) - ln (7)

= \frac{1}{2} (ln (63) - ln (7))

Simplify

= ln (3)

\int_{1}^{4} (2 x - 2 x) d x

f (x) = 7 + 4 x^{2} - 2 x^{3}, (1, 9)

\int \frac{2 x}{- 5 - 6 x ^{2} - x ^{4}} d x

\frac{7 x}{8 - tan ( x )}

\int \frac{4 x ^{2} + x + 4}{( x ^{2} + 1 ) ^{2}} d x

What is the integral from 2 to 9 of 4/(8x-9) ?
The integral from 2 to 9 of 4/(8x-9) is ln(3)